Question

11. A roller 1.5 m long has a diameter of 70 cm. How many revolution will it make to level a
playground measuring 50 m x 33 m.
12. The diameter of a roller is 42 cm and its length is 120 cm. If takes 500 complete revolution to
move once to land a playground, find the area of the playground in m².​

Answer 1

Answer:

11.length of roller or height = 1.5 m diameter of roller =0.7 m radius of roller = 0.7/2 m

outer area of cylinder = 2nrh

= 2 x 22/7 x 0.7/2 x 1.5

= 3.3 m sq

area of field = 50x33 = 1650 m sq

no. of revolution taken = 1650/3.3 = 500 revolutions

12.We have the diameter of a cyclindrial roller = 42 cm

⇒ The radius of cyclindrical roller (r) = 42 / 2 = 21 cm

Length of a cyclindrical roller (h) = 120 cm

Curved surface of the roller = 2πrh = 2 × 22 / 7 × 21 × 120 = 15840 cm2

= 15840 / 10000 m2 = 1.584 m2

Area covered by the roller in one complete revolution .

= curved surface of the roller = 1.584 m2

∴ Area of the playground = Area covered by the roller in 500 complete revolutions

= 500 × 1.584 = 792 m2

Step-by-step explanation:

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Answer 2

11 number answer is above in the attachment

12) Radius of the roller (r)=84/2cm=42cm

length of the roller (h)=120cm

∴ Area of the playground levelled in taking 1 complete revolution 2πrh

=2×227×42×120=31680cm^2

∴ Area of the playground

=31680×500=15840000cm^2

15840000/100×100 m^2=1584m^2